Speed of Wave Problem

  • #1
bobacity
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Homework Statement:
In a uniform frequency and amplitude sound wave, a single air molecule vibrates with a period of 0.025s, and across a width of 3m. What is the speed of this wave?
Relevant Equations:
wavelength = velocity/frequency
f = 1/T
Since the period is 0.025s, I think the frequency is 1/0.025s = 40 Hz. I don't know how can I proceed solving the problem from here. I'm assuming I will have to try to find the velocity and wavelength, but idk how.
 

Answers and Replies

  • #2
TSny
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The wording of the question seems strange. It appears to imply that a single air molecule oscillates over a distance of 3 m.
 
  • #3
kuruman
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The wording of the question seems strange. It appears to imply that a single air molecule oscillates over a distance of 3 m.
Yes, that is a room-sized distance. Even if the 3 m amplitude is a typo, I don't see how it can be related to the wavelength because this sound wave has "uniform frequency and amplitude" and therefore can be assumed to be harmonic. The only alternative interpretation that I can think of is that the 3 m is the distance between occurrences of the same phase of oscillating molecules, i.e. the relative velocity between molecules 3 m apart is zero. I am grasping at straws here.
 
  • #4
BvU
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Could it be a standing wave with nodes every 3 m?

:wink:

##\ ##
 
  • #5
kuruman
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Could it be a standing wave with nodes every 3 m?

:wink:

##\ ##
I thought about standing waves, but then what is the 3 m all about? The problem clearly states, "a single air molecule vibrates with a period of 0.025s, and across a width of 3m" which can only be interpreted that an air molecule vibrates with amplitude 1.5 m. If the nodes are 3 m apart, that is not the distance traveled by the vibrating air molecules (see video simulation below) but maybe the author of the problem thought that they do.

 
  • #6
jbriggs444
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My guess is that the author labors under the idea that the molecules in a compression wave oscillate like a sine wave. So the wavelength must be ##\pi## times the peak to trough displacement. [Of course, the peak to trough displacement actually depends on the amplitude and is independent of the wavelength]

With this assumption in mind, we could arrive at a calculated speed of sound that is a bit high, but not outrageously so.
 
  • #7
kuruman
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My guess is that the author labors under the idea that the molecules in a compression wave oscillate like a sine wave. So the wavelength must be ##\pi## times the peak to trough displacement. [Of course, the peak to trough displacement actually depends on the amplitude and is independent of the wavelength]

With this assumption in mind, we could arrive at a calculated speed of sound that is a bit high, but not outrageously so.
Please help me understand your guess about the author's model. We have a harmonic wave that we can write as $$y(x,t)=A~\sin\left[2\pi(\frac{x}{\lambda}-\frac{t}{T})\right]$$We are looking for the speed of propagation ##\frac{\lambda}{T}##. The period ##T## is given. How do we extract ##\lambda## and what is its relation to the given 3 m? If I take a snapshot of the traveling wave (below), the peak-to-trough distance is indicated in red. Clearly, any given molecule oscillates with amplitude ##A## that is a fraction of that. Also, I don't understand how the wavelength can be ##\pi\frac{\lambda}{2}.## I suppose I don't understand what the peak-to-trough distance ought to be according to the guessed at author's model.

LongWave.png
 
  • #8
jbriggs444
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Please help me understand your guess about the author's model.

The theory is that the author is thinking of graphs of the form ##a \sin \frac{x}{a}## so that the slope of the graph at ##x=0## is always 1 -- a classic sine wave scaled evenly in both dimensions.

Edit: Oh, I see the disconnect. The "peak to trough" measure is where we are failing to understand one another. By this I meant the vertical span from the peak of the sine wave (##\sin \frac \pi 2##) to the trough of the sine wave (##\sin \frac {3 \pi} 2##). This span is 2. Meanwhile, the wavelength for one cycle is ##2 \pi##. The ratio between the two is just ##\pi##.

If we have an air molecule oscillating with a span of 3 meters then the [mistaken] wavelength is ##3\pi## meters. We can multiply by frequency and obtain a speed. The virtue of the theory is that the resulting speed is something like 10% higher than the actual speed of sound in room temperature air if my calculations are correct. To make the numbers actually work, it would have to be a very hot day or an atmosphere polluted with a significant amount of helium.

Obviously this is a mistake. The graph of a sine wave can scale differently in the two dimensions. Wavelength and amplitude are orthogonal and need not even share the same units.
 
Last edited:
  • #9
kuruman
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I see. Thanks for the clarification.
 

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